We started the day getting to know the physical form of inductor. There are two metal tips on the corner of the inductor to be connected to the power supply, so the current can flow through. We learned from past lecture that inductor stores the current, just like capacitor stores voltage. We also learned that inductor is flux per current. Today, we predict the inductor vs time graph. We know that voltage vs time of inductor is a 1/ln graph. So we found that inductor vs time is an exponential graph.
We first are given a question, with known as below. Number of coils is 110, length of 0.05m, radius of 5.12 * 10^-4m. First, we calculate the area of the 18 gauge copper coil using the area of circle formula. We know the density of copper to be 1.72 * 10^-8. By plugging in the numbers for density multiplied by length per area, we can find the resistance to be 0.3 ohm. We then calculate the inductance using formula of inductance from past lecture to get the number of coils, which is inductance = myu*N(square)*A/l. For this, we use are of the inductor, which is a square and we also use the magnetic permeability of free space, which we know to be 4pi*10^-7. Plugging in the knowns, we found the inductance to be 760*10^-6 H. We then can calculate the tao by inductance divided by resistance, which we found to be 5.07*10^-6.
We are then given a 880 coils inductor. By connecting it to the oscilloscope and the power generator using the alligator cables, we firstly set the function generator to a square wave form. Then we adjust position, and vertical and horizontal scales. Lastly, we adjust frequency of the function generator. We find the voltage vs time graph to be as below:
From the graph, we obtained the half- time to be 0.4 microsecond. We then calculate the tao by half-time divided by ln 2. Then we can also calculate the inductance by multiplying the resistance and tao. Lastly, we can use the formula of inductance from past lecture to get the number of coils, which is inductance = myu*N(square)*A/l. There is an error in the inductor's calibration.
Using the last color of the resistor, which is brown. We can find the uncertainty for the resistor in the table as we learned before, which is 5% +/- 2*10^-7
We then given a circuit with an inductor, resistors, and voltage as in orange diagram below. First, we want to calculate Tao by dividing the inductors by the resistor, both in line 1. After finding that, we can calculate the half time, which is 5* the Tao that we found. Now we want to find current of each line. For first current, we will use the equation Max current multiplied by 1-e^-t/Tao. We found the Tao earlier and we can use the time as 170 ms. Max current is the voltage in the circuit divided by the resistor in first line. We found the answer to be 0.165 A. For the second current, we can just calculate the current using the voltage divided by the resistor in the second line and we found the answer to be 0.0616 A.
Next, we want to find the potential drop across the resistor. We know that the voltage in the second line is just the 45V. On the first line, we can calculate the voltage by multiplying the current that we found by the resistor on the first line and we found the voltage to be 19.8. When we subtract this we found the potential drop to be 25 V.
We can also calculate the time for inductor's voltage to drop to 11V. By subtracting the initial voltage to the final voltage. Then divide them by the first line's resistor to find the current on that line. Then we can find the time by using the current equation earlier. By using current max and Tao on line one, we found the time to be 409ms.
We can also find the energy dissipated after switch is open for 170ms. By using the equation of emf energy from past lecture, we can calculate it by multiplying half of current(square) by inductance on the first line. We find it to be 4.76*10^-4 J.
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